If $x \bigtriangledown y = x^{2}-4y^{2}$ and $x \boxdot y = 2x-6$, find $(0 \boxdot -4) \bigtriangledown 1$.
Solution: First, find $0 \boxdot -4$ $ 0 \boxdot -4 = (2)(0)-6$ $ \hphantom{0 \boxdot -4} = -6$ Now, find $-6 \bigtriangledown 1$ $ -6 \bigtriangledown 1 = (-6)^{2}-4(1^{2})$ $ \hphantom{-6 \bigtriangledown 1} = 32$.